TITLE: Is the complement of a C7 graph planar or non-planar?
QUESTION [5 upvotes]: I am trying to prove whether the complement of a cycle-7 graph is planar or non-planar.
I tried to use Euler's theorem to prove it;
In a connected simple planar graph with v vertices and e edges, if v ≥ 3, then e ≤ 3v−6.
I have that the complement of a C7 has 7 vertices and 14 edges.. Plugging this into Euler's theorem this comes out as 14 ≤ 15, which obviously holds, so this method does not prove non-planarity.
I've also tried to find K5 and a K3,3 as subgraphs but also had no luck there either. 
Any helpful tips on how to go about it from here?
I've included a photo of the only version of the complement of C7 I could find, apologies for the vertices being labelled as weekdays. I just wanted to show the graph more than anything

REPLY [2 votes]: While the question is perfectly answered by citing the Kuratowski theorem, it is amusing to note that this workday graph actually defines a Moebius strip. You can build it with paper consisting of 7 triangles. The fact that it is not planar can be seen now as a consequence that it is not possible to ``flatten" the Moebius strip: it would provide an orientation, but the Moebius strip is non-orientable. Picture of Workday graph as Moebius strip